Approximaths

Perturbation Theory in Action

We would like to use perturbation methods similar to those presented for solving polynomials in the previous posts, but this time to solve differential equations. We will solve the following basic first-order equation with the corresponding initial condition:

$\displaystyle y' = y,\; y(0)=1$

This equation can be solved exactly very easily. The answer is $y(x)=e^x$ . Now, let us pretend that we don’t know how to solve this equation exactly and use perturbation theory techniques. We write the function $y(x)$ in the form (in a similar way to how we calculated the roots of polynomials):

$\displaystyle y_{\epsilon}(x)=\sum_{n=0}^{\infty}a_n(x)\epsilon^n=a_0(x)+a_1(x)\epsilon+a_2(x)\epsilon^2+\ldots$

According to the definition above, we have:

$\displaystyle y_{\epsilon}(x)=a_0(x)+a_1(x)\epsilon+a_2(x)\epsilon^2+a_3(x)\epsilon^3+\ldots$

$\displaystyle y'_{\epsilon}(x)=a_0'(x)+a_1'(x)\epsilon+a_2'(x)\epsilon^2+a_3'(x)\epsilon^3+\ldots$

Now we can perturb the differential equation above:

$\displaystyle y'=\epsilon y$

and write (up to terms of degree 3):

$\displaystyle a_0'(x)+a_1'(x)\epsilon+a_2'(x)\epsilon^2+a_3'(x)\epsilon^3=\epsilon(a_0(x)+a_1(x)\epsilon+a_2(x)\epsilon^2)$

$\displaystyle a_0'(x)+a_1'(x)\epsilon+a_2'(x)\epsilon^2+a_3'(x)\epsilon^3-a_0(x)\epsilon-a_1(x)\epsilon^2-a_2(x)\epsilon^3=0$

$\displaystyle a_0'(x)+(a_1'(x)-a_0(x))\epsilon+(a_2'(x)-a_1(x))\epsilon^2+(a_3'(x)-a_2(x))\epsilon^3=0$

Therefore, we have to solve the following system of equations:

$\displaystyle a_0'(x)=0$

$\displaystyle a_1'(x)-a_0(x)=0$

$\displaystyle a_2'(x)-a_1(x)=0$

$\displaystyle a_3'(x)-a_2(x)=0$

Applying the initial conditions $a_0(0)=1$ and $a_n(0)=0$ for $n\ge1$ , we obtain:

$\displaystyle a_1'(x)-1=0\implies a_1=x$

$\displaystyle a_2'(x)-a_1(x)=0\implies a_2=\frac{1}{2}x^2$

$\displaystyle a_3'(x)-a_2(x)=0\implies a_3=\frac{1}{3!}x^3$

Therefore:

$\displaystyle y_{\epsilon}(x)=1+x\epsilon+\frac{1}{2}x^2\epsilon^2+\frac{1}{3!}x^3\epsilon^3$

and $y(x)$ is approximated as follows (setting $\epsilon=1$ ):

$\displaystyle y_{\epsilon}(x)=1+x+\frac{1}{2}x^2+\frac{1}{3!}x^3$

which corresponds to the first terms of the Taylor series of $e^x$ (the exact solution).

July 30, 2026
The Newton Polygon

The construction based on the affine exponent functions (see this post)

$\displaystyle L_i(p)=m_ip+\alpha_i$

has a simple geometric counterpart known as the Newton polygon. Instead of representing each monomial

$\displaystyle \epsilon^{\alpha_i}x^{m_i}$

by the affine function $L_i(p)$ , we associate it with the point

$\displaystyle (m_i,\alpha_i)$

in the $(m,\alpha)$ plane. The Newton polygon is defined as the lower convex hull of these points. Its edges contain exactly the information needed to determine the admissible scalings. Indeed, if an edge joins the points

$\displaystyle (m_i,\alpha_i) \qquad\text{and}\qquad (m_j,\alpha_j)$

its slope is

$\displaystyle s=\frac{\alpha_j-\alpha_i}{m_j-m_i}$

and the corresponding scaling exponent is

$\displaystyle p=-s=-\frac{\alpha_j-\alpha_i}{m_j-m_i}$

Thus, each edge of the Newton polygon corresponds to a dominant balance between monomials. The successive edges recover exactly the same admissible scalings as the corners of the lower envelope of the affine functions $L_i(p)$ . The Newton polygon therefore provides a dual geometric interpretation of the regularization process. As an illustration, consider the perturbed polynomial (see this post)

$\displaystyle P_{\epsilon}(x)=\epsilon^3x^3+x^2+\epsilon x$

The corresponding points in the $(m,\alpha)$ plane are

$\displaystyle (3,3),\qquad (2,0),\qquad (1,1)$

The Newton polygon is the lower convex hull (see figure below)

$\displaystyle (1,1)\longrightarrow(2,0)\longrightarrow(3,3)$

The first edge has slope

$\displaystyle s=\frac{0-1}{2-1}=-1$

and therefore gives the scaling

$\displaystyle p=1$

The second edge has slope

$\displaystyle s=\frac{3-0}{3-2}=3$

and gives the scaling

$\displaystyle p=-3$

These two values are exactly the admissible scalings obtained from the intersections of the affine exponent functions

$\displaystyle 3p+3,\qquad 2p,\qquad p+1$

The left panel of the figure below shows the affine exponent functions and their lower envelope, while the right panel shows the corresponding Newton polygon. The two pictures are dual representations of the same dominant balance structure.

July 23, 2026
Regularize V

The previous post illustrates the general strategy: the admissible scalings are obtained from the intersections of the affine exponent functions. In this post, we explain why this construction works. Consider a perturbed polynomial of the form

$\displaystyle P_{\epsilon}(x)=\sum_{i=1}^{n}\epsilon^{\alpha_i}x^{m_i}$

After the scaling

$\displaystyle x=\epsilon^{p}w$

each monomial becomes

$\displaystyle \epsilon^{\alpha_i}(\epsilon^p w)^{m_i} = \epsilon^{\alpha_i+m_ip}w^{m_i}$

Thus every monomial defines an affine function

$\displaystyle L_i(p)=m_ip+\alpha_i$

The exponents of $\epsilon$ are therefore completely described by the family of affine functions

$\displaystyle L_1(p),L_2(p),\ldots,L_n(p)$

The key observation is that, as $\epsilon\to0$ , the smallest exponent dominates all the others. Indeed, if

$\displaystyle \alpha<\beta$

then

$\displaystyle \epsilon^\alpha\gg\epsilon^\beta$

Consequently, for a fixed value of $p$ , only the monomials whose exponent is minimal contribute to the dominant part of the polynomial.

Now suppose that one affine function lies strictly below all the others. Then a single monomial dominates, and the limit polynomial consists of only one term.

The interesting situation occurs when two affine functions intersect and their common value is the minimum among all the exponents. At such a point,

$\displaystyle L_i(p)=L_j(p)=\min_k L_k(p)$

so two monomials have exactly the same asymptotic order. After normalization, both survive in the limit polynomial, producing a non-trivial dominant part.

Not every intersection is relevant. Two affine functions may intersect while another affine function remains strictly below them. In that case, the common exponent is not minimal, and the corresponding scaling does not contribute to the dominant polynomial.

Therefore, the admissible scalings are obtained precisely from the intersection points where the common exponent is minimal.

Geometrically, this means that the admissible scalings correspond to the corners of the lower envelope of the affine functions (see previous post). This simple geometric picture explains why the graph of the exponents completely determines the regularization process.

July 16, 2026
Regularize IV (scaling)

Let us consider the following perturbed polynomial:

$\displaystyle P(x)=x^3+x^2+x$

$\displaystyle P_{\epsilon}(x)=\epsilon^3x^3+x^2+\epsilon x$

$\displaystyle P_{\epsilon}(\epsilon^p w)=\epsilon^3(\epsilon^p w)^3+(\epsilon^p w)^2+\epsilon(\epsilon^p w)$

$\displaystyle =\epsilon^{3p+3}w^3+\epsilon^{2p}w^2+\epsilon^{p+1}w$

In the figure below, we plot the graphs of the affine functions $3p+3$ , $2p$ , and $p+1$ . Their intersection points determine the values of $p$ for which two exponents coincide. Among these, we retain only the intersections where the common exponent is minimal. In this example, we obtain

$\displaystyle (p_j,\nu_j)=(-3,-6)$

We now compute $T_{\epsilon}^{(j)}(w)$ defined as follows:

$\displaystyle T_{\epsilon}^{(j)}(w):=\epsilon^{-\nu_j}P_{\epsilon}(\epsilon^{p_j}w)$

$\displaystyle T_{\epsilon}^{(j)}(w)=\epsilon^{6}P_{\epsilon}(\epsilon^{-3}w)$

$\displaystyle =\epsilon^{6}\bigl(\epsilon^{3(-3)+3}w^3+\epsilon^{2(-3)}w^2+\epsilon^{-3+1}w\bigr)$

$\displaystyle =\epsilon^{6}\bigl(\epsilon^{-6}w^3+\epsilon^{-6}w^2+\epsilon^{-2}w\bigr)$

$\displaystyle =\underbrace{w^3+w^2}_{T_{0}^{(j)}(w)}+\underbrace{\epsilon^{4}w}_{E_{\epsilon}^{(j)}(w)}$

Observe that $E_{0}^{(j)}(w)=0$ . Therefore, we may write

$\displaystyle T_{\epsilon}^{(j)}(w)=T_{0}^{(j)}(w)+E_{\epsilon}^{(j)}(w)$

The polynomial $T_{0}^{(j)}(w)$ is the dominant part of $T_{\epsilon}^{(j)}(w)$ as $\epsilon\to0$ , since it is independent of $\epsilon$ . In this example, the dominant polynomial is non-degenerate. Therefore, the scaling $x=\epsilon^{p_j}w$ regularizes the perturbed polynomial $P_{\epsilon}(x)$ .

Graphs of the affine functions $3p+3$ , $2p$ , and $p+1$ . The point $(-3,-6)$ is the intersection of the lines $3p+3$ and $2p$ . At $p=-3$ , the minimum exponent is attained simultaneously by two monomials.

July 9, 2026
Regularization III

In this post, we justify the regularization procedure introduced in previous posts for singular polynomial problems. Consider:

$\displaystyle P(x)=a_0+a_1x+\dots+a_nx^n$

$\displaystyle P(x)=1+A_1x+\dots+A_nx^n$

$\displaystyle P_{\epsilon}(x)=f_0(\epsilon)+f_1(\epsilon)\epsilon^{\alpha_1}A_1x+\dots+f_n(\epsilon)\epsilon^{\alpha_n}A_nx^n$

$\displaystyle P_{\epsilon}(x)=(1+b_0\epsilon+c_0\epsilon^2+\dots)+(1+b_1\epsilon+c_1\epsilon^2+\dots)\epsilon^{\alpha_1}A_1x+\dots+(1+b_n\epsilon+c_n\epsilon^2+\dots)\epsilon^{\alpha_n}A_nx^n$

where $f_0(\epsilon),f_1(\epsilon),\dots,f_n(\epsilon)$ admit power-series expansions in $\epsilon$ , $\alpha_i$ ‘s are rational numbers and $b_i$ ‘s, $c_i$ ‘s are constants.

The last expression is therefore a generalized version of a perturbed polynomial

. We can rearrange this expression as follows:

$\displaystyle P_{\epsilon}(x)=(1+b_0\epsilon+c_0\epsilon^2+\dots)$

$\displaystyle +(1+b_1\epsilon+c_1\epsilon^2+\dots)\epsilon^{\alpha_1}A_1x$

$\displaystyle +\dots$

$\displaystyle +(1+b_n\epsilon+c_n\epsilon^2+\dots)\epsilon^{\alpha_n}A_nx^n$

Collecting terms according to their powers of $\epsilon$ , we obtain:

$\displaystyle P_{\epsilon}(x)=1+\epsilon^{\alpha_1}A_1x+\dots+\epsilon^{\alpha_n}A_nx^n$

$\displaystyle +b_0\epsilon+b_1\epsilon^{\alpha_1+1}A_1x+\dots+b_n\epsilon^{\alpha_n+1}A_nx^n$

$\displaystyle +\dots$

Theorem (Newton–Puiseux). Each root of the polynomial above is of the form:

$\displaystyle x(\epsilon)=\epsilon^pw(\epsilon),\quad w(0)\neq0$

where $w(\epsilon)$ is a continuous function near zero. For convenience, we simply write $x=\epsilon^pw$ . If the theorem holds, the polynomial above becomes:

$\displaystyle P_{\epsilon}(\epsilon^pw)=1+\epsilon^{\alpha_1}A_1\epsilon^pw+\dots+\epsilon^{\alpha_n}A_n(\epsilon^pw)^n$

$\displaystyle +b_0\epsilon+b_1\epsilon^{\alpha_1+1}A_1\epsilon^pw+\dots+b_n\epsilon^{\alpha_n+1}A_n(\epsilon^pw)^n$

$\displaystyle +\dots$

We define $Q_{\epsilon}(w):=1+\epsilon^{\alpha_1+p}A_1w+\dots+\epsilon^{\alpha_n+np}A_nw^n$ and have:

$\displaystyle P_{\epsilon}(\epsilon^pw)=Q_{\epsilon}(w)+\epsilon(b_0+b_1\epsilon^{\alpha_1+p}A_1w+\dots+b_n\epsilon^{\alpha_n+np}A_nw^n)+\dots$

If $\epsilon^pw$ is a root of $P_{\epsilon}(x)$ , we must satisfy:

$\displaystyle \lim_{\epsilon\to0}P_{\epsilon}(\epsilon^pw)=\lim_{\epsilon\to0}Q_{\epsilon}(w)=0$

Since $w(\epsilon)$ is a continuous function near zero, we inspect the structure of $Q_{\epsilon}(w(0))$ to determine the leading-order behavior:

$\displaystyle Q_{\epsilon}(w(0))=1+\epsilon^{\alpha_1+p}A_1w(0)+\dots+\epsilon^{\alpha_n+np}A_nw(0)^n$

Considering the set of its exponents of $\epsilon$ :

$\displaystyle E=\{0,\alpha_1+p,\dots,\alpha_k+kp,\dots,\alpha_n+np\}$

In order for the condition $\lim_{\epsilon\to0}Q_{\epsilon}(w)=0$ to hold, the minimal value of the set $E$ must be exactly $0$ , and this minimum must be shared by at least two distinct exponents. These identical minimal exponents define the dominant terms of $Q_{\epsilon}(w(0))$ which can then compensate each other to balance the equation at leading order.

July 2, 2026
Regularize II

$\displaystyle$

The underlying idea of regularization is to recover the roots that are lost in a singular limit. When some coefficients of a polynomial are multiplied by powers of a small parameter $\epsilon$ , the limiting polynomial obtained by setting $\epsilon=0$ may have fewer roots than the original polynomial. In this situation, some roots have escaped to a different scale and are therefore invisible in the limit.

To recover these missing roots, we introduce a rescaling

$\displaystyle x:=\epsilon^p w$

where the exponent $p$ is chosen so that at least two dominant terms of the polynomial balance each other. Geometrically, this choice is determined from the Newton polygon (or Newton diagram) by examining the exponents of $\epsilon$ . After an additional normalization by a factor $\epsilon^{-m_j}$ , one obtains a new limiting polynomial in the variable $w$ . This reduced polynomial captures the asymptotic behavior of the roots on the corresponding scale and typically restores the correct number of roots.

In this sense, the singularity is not an intrinsic property of the problem but rather the consequence of observing the problem at an inappropriate scale.

As a simple example, consider

$\displaystyle P(x) = x^2 -x+1$

$\displaystyle P_\epsilon(x) =\epsilon x^2-x+1$

Setting $\epsilon=0$ gives

$\displaystyle P_0(x)=-x+1$

which has only one root, although the original polynomial is quadratic and therefore has two roots. To recover the missing root, let

$\displaystyle x:=\epsilon^{-1}w$

Substituting into the polynomial gives

$\displaystyle P_\epsilon(\epsilon^{-1}w) = \epsilon(\epsilon^{-2}w^2)-\epsilon^{-1}w+1 = \epsilon^{-1}(w^2-w+\epsilon)$

Multiplying by $\epsilon$ yields

$\displaystyle w^2-w+\epsilon$

Taking the limit $\epsilon\to0$ gives the regular polynomial

$\displaystyle w^2-w=w(w-1)$

which has two roots. One corresponds to the finite root seen in the original limit, while the second reveals a root located on the larger scale $x=O(\epsilon^{-1})$ . The rescaling has therefore recovered the root that was hidden in the singular limit.

June 25, 2026
Regularize I

We will now present (without justification) a method for regularizing a singular problem involving a polynomial. We want to find the roots of :

$\displaystyle P(x) = x^6 - x^4 - x^3 + 8$

We decide to perturb the above polynomial as follows:

$\displaystyle P_{\epsilon}(x) = \epsilon^2 x^6 - \epsilon x^4 - x^3 + 8$

For $\epsilon = 0$ we have:

$\displaystyle P_{\epsilon= 0}(x) = -x^3 + 8$

This polynomial has three roots. We have a singular problem since the original polynomial has six roots.

We can try to fix the problem using a change of variable:

$\displaystyle x := \epsilon^p w$

$\displaystyle P_{\epsilon}(\epsilon^p w) = \epsilon^2 (\epsilon^p w)^6 - \epsilon (\epsilon^p w)^4 - (\epsilon^p w)^3 + 8$

$\displaystyle = \epsilon^2 (\epsilon^{6p} w^6) - \epsilon (\epsilon^{4p} w^4) - (\epsilon^{3p} w^3) + 8$

$\displaystyle = \epsilon^{6p+2} w^6 - \epsilon^{4p+1} w^4 - \epsilon^{3p} w^3 + 8$

Now we collect the exponents of $\epsilon$ : ( $6p+2, 4p+1$ , $3p, 0$ ) and plot them in a graph as a function of p (see figure below). If we think of the lines on the graph as delimiting a figure in the plane, we can select the vertex of this figure with the largest p value and the smallest f(p) value. In our case, the point corresponding to this criterion is the point of intersection of the lines $6p+2$ and $3p$ . We thus obtain the point of intersection with coordinates ( $-2/3, -2$ ).

We will call the coordinates of the intersection point $p_j$ and $m_j$ respectively.

Now calculate:

$\displaystyle \epsilon^{-m_j} P_{\epsilon}(\epsilon^{p_j} w) = \epsilon^{2} P_{\epsilon}(\epsilon^{-\frac{2}{3}} w)$

$\displaystyle = \epsilon^2 (\epsilon^{6(-\frac{2}{3})+2} w^6 - \epsilon^{4(-\frac{2}{3})+1} w^4 - \epsilon^{3(-\frac{2}{3})} w^3 + 8)$

$\displaystyle = \epsilon^2 (\epsilon^{-2} w^6 - \epsilon^{-\frac{5}{3}} w^4 - \epsilon^{-2} w^3 + 8)$

$\displaystyle = w^6 - \epsilon^{\frac{1}{3}} w^4 - w^3 + 8 \epsilon^2$

This is now a regular problem since the new polynomial:

$\displaystyle P(w) = w^6 - w^3$

has 6 roots.

Using the procedure above we have regularized a singular problem involving a polynomial using the following sequence:

$\displaystyle P(x) \to P_{\epsilon}(x) \to P_{\epsilon}(\epsilon^p w) \to \epsilon^{-m_j} P_{\epsilon}(\epsilon^{p_j} w)$

June 18, 2026
A quintic equation
So far, we’ve introduced the perturbation technique, described regular and irregular problems and presented solutions to simple problems that we could have solved exactly without using perturbative series. We will now consider a problem that cannot be solved exactly. In this way, we hope to illustrate the power of this method.

We want to find a real zero of the following quintic equation:

$\displaystyle x^5 + x = 1$

We will proceed as usual (see previous sections):

$\displaystyle x^5 + \epsilon x = 1$

$\displaystyle (a_0 + a_1\epsilon + a_2 \epsilon^2 + \cdots)^5 + \epsilon (a_0 + a_1\epsilon + a_2 \epsilon^2 + \cdots) = 1$

Solving the unperturbed problem (where $\epsilon = 0$ ) we see that $a_0 = 1$ ( $a_0$ is always the answer to the unperturbed problem). We can use Python to expand the perturbative series and compute the coefficients (see code below). In this example we will calculate up to six coefficients.
```
from sympy import *

# set variables
a1 = Symbol('a1')
a2 = Symbol('a2')
a3 = Symbol('a3')
a4 = Symbol('a4')
a5 = Symbol('a5')
x = Symbol('x') # epsilon

# from the unperturbed problem we know that a0 = 1

# expand the quintic equation
equ = expand ((1 + a1*x + a2*x**2 + a3*x**3 + a4*x**4 + a5*x**5)** 5 + x*(1 + a1*x + a2*x**2 + a3*x**3 + a4*x**4 + a5*x**5))

# get the coefficients separately
equ.coeff(x)

5*a1+1

equ.coeff(x**2)

10*a1**2+a1+5*a2

# solve the system of equations to calculate coefficients
solution = solve([equ.coeff(x),equ.coeff(x**2),equ.coeff(x**3)], (a1, a2, a3))
solution

[(-1/5, -1/25, -1/125)]

solution = solve([equ.coeff(x),equ.coeff(x**2),equ.coeff(x**3),equ.coeff(x**4), equ.coeff(x**5)], (a1, a2, a3, a4, a5))
solution

[(-1/5, -1/25, -1/125, 0, 21/15625)])
```
Therefore (using the results above calculated using Python):

$\displaystyle a_0 = 1$

$\displaystyle a_1 = -\frac{1}{5}$

$\displaystyle a_2 = -\frac{1}{25}$

$\displaystyle a_3 = -\frac{1}{125}$

$\displaystyle a_4 = 0$

$\displaystyle a_5 = \frac{21}{15625}$

We have:

$\displaystyle x(\epsilon) = 1 - \frac{1}{5}\epsilon - \frac{1}{25}\epsilon^2 - \frac{1}{125}\epsilon^3 + 0\cdot\epsilon^4 + \frac{21}{15625}\epsilon^5 + \cdots$

An approximation (fifth-order approximation) of the real solution of $x^5 + x = 1$ is (setting $\epsilon = 1$ ):

$\displaystyle 1 - \frac{1}{5} - \frac{1}{25} - \frac{1}{125} + 0 + \frac{21}{15625} \approx 0.7533$
June 11, 2026
Illustration of a singular problem

Illustration of a singular problem:

$\displaystyle x^2-2x +1 = 0 \quad \text{Problem}$

$\displaystyle \epsilon x^2-2x +1 = 0 \quad \text{Insert }\epsilon$

The solutions to the perturbed problem are (see figure below):

$\displaystyle x_1(\epsilon) = \frac{1+ \sqrt{1-\epsilon}}{\epsilon}$

$\displaystyle x_2(\epsilon) = \frac{1- \sqrt{1-\epsilon}}{\epsilon}$

While one of the roots approaches $\frac{1}{2}$ as $\epsilon \to 0$ , the other goes to infinity. This is a manifestation of a singular behaviour. The problems above illustrate the importance of setting $\epsilon$ to a right position.

A solution of the regular problem $x^2-2x +\epsilon = 0$ is $x_1(\epsilon) = 1 - \sqrt{1- \epsilon}$ . We would like to calculate the corresponding perturbative series.
Using:

$\displaystyle (a+b)^n = \sum_{k=0}^{\infty} {n\choose k} a^{n-k}b^{k}$

$\displaystyle = a^n + n a^{n-1}b + \frac{n(n-1)}{2} a^{n-2}b^{2}+...+ \frac{n(n-1)\cdots(n-k+1)}{k!} a^{n-k}b^k +...$

We would like to find the perturbative series corresponding to:

$\displaystyle \sqrt{1-\epsilon} = (1-\epsilon)^{1/2}$

with $a=1$ , $b = -\epsilon$ , $n = \frac{1}{2}$ therefore:

$\displaystyle \sqrt{1-\epsilon} = 1 - \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 - ...$

$\displaystyle 1- \sqrt{1-\epsilon} = \frac{1}{2}\epsilon + \frac{1}{8}\epsilon^2 + ...$

Here are the first series and their corresponding graphs:

$\displaystyle s_2 = \frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2$

$\displaystyle s_3 =\frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2 + \frac{1}{16} \epsilon^3$

$\displaystyle s_4 = \frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2 + \frac{1}{16} \epsilon^3 + \frac{5}{128} \epsilon^4$

June 4, 2026
Regular Problem II

Illustration of another regular problem:

$\displaystyle x^2-2x -1 = 0 \quad \text{Problem}$

$\displaystyle x^2-2\epsilon x-1=0 \quad \text{Insert }\epsilon$

The solutions to the perturbed problem are (see figure below):

$\displaystyle x_1(\epsilon) = \frac{2\epsilon + \sqrt{4 \epsilon^2 + 4}}{2} = \epsilon + \sqrt{1 + \epsilon^2}$

$\displaystyle x_2(\epsilon) = \frac{2\epsilon - \sqrt{4 \epsilon^2 + 4}}{2} = \epsilon - \sqrt{1 + \epsilon^2}$

The figure above shows solutions of the problem $x^2-2\epsilon x-1 = 0$ using perturbation theory. For $\epsilon =0$ the solutions are -1 and 1 and for $\epsilon =1$ the solutions are exactly $1+\sqrt{2}$ and $1-\sqrt{2}$ .

$\displaystyle x_1(\epsilon) = 1 + \epsilon + \frac{1}{2}\epsilon^2 - \frac{1}{8}\epsilon^4 + \frac{1}{16}\epsilon^6 - \frac{5}{128}\epsilon^8 + O(\epsilon^{10})$

$\displaystyle x_2(\epsilon) = -1 + \epsilon - \frac{1}{2}\epsilon^2 + \frac{1}{8}\epsilon^4 - \frac{1}{16}\epsilon^6 + \frac{5}{128}\epsilon^8 + O(\epsilon^{10})$

Setting $\epsilon = 1$ :

$\displaystyle x_1(1) \approx 2 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \frac{5}{128} = 2.3984375$

$\displaystyle x_2(1) \approx 0 - \frac{1}{2} + \frac{1}{8} - \frac{1}{16} + \frac{5}{128} = -0.3984375$

This simple example clearly illustrates the application of regular perturbation theory to a quadratic equation, showing how the solutions vary continuously with the perturbation parameter $\epsilon$ .

May 28, 2026